Question
Download Solution PDFIf the temperature of the heat source is 1990 K and the sink is 850 K, then what is the Carnot efficiency
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Carnot Efficiency:
Carnot Heat engine is a reversible ideal heat engine and the efficiency of the reversible engine is
\(\eta = \frac{{Net\;work\;output\;to\;the\;cycle}}{{Total\;heat\;input\;to\;the\;cycle}} = \frac{{{W_{net}}}}{{{Q_H}}}\)
Where,
Wnet = Network done by cycle = QH – QL
QH = Heat input to the cycle from source, QL = Heat rejected to sink
TH = Source temperature, TL = Sink temperature
\(\eta = \frac{{{W_{net}}}}{{{Q_H}}} = \frac{{{Q_H} - {Q_L}}}{{{Q_H}}} = 1 - \frac{{{Q_L}}}{{{Q_H}}}\)
For reversible heat engine ⇒ \(\frac{{{Q_H}}}{{{Q_L}}} = \frac{{{T_H}}}{{{T_L}}}\)
\(\eta = 1 - \frac{{{Q_L}}}{{{Q_H}}} = 1 - \;\frac{{{T_L}}}{{{T_H}}}\)
Calculation:
Given:
The temperature of heat source, TH = 1990 K, The temperature of the heat sink, TL = 850 K
Carnot efficiency ⇒ \(\eta = 1 - \;\frac{{{T_L}}}{{{T_H}}} = \;1 - \;\frac{{850}}{{1990}} = 0.573\)
η = 0.573 or 57.3 %
Last updated on May 28, 2025
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